0195136047.pdf

120 TIME-DEPENDENT CIRCUIT ANALYSIS

(a) Since the Thévenin impedance is the ratio of the open-circuit voltage to the short-circuit

current, calculation with respect to terminalsA–Bis shown next. Let us first calculate the

open-circuit voltageV ̄oc.

The KVL equation for the left-hand loop is

( 200 +j 20 )I ̄= 20 0°, or I ̄=

20  0°

200 +j 20

Then,

V ̄oc=j 20 I−

(

−j 100 I ̄

)

=j 120 I ̄=

(j^120 )(^20 )

200 +j 20

= 1. 19 +j 11. 88

= 11. 94  84 .3° V

The short-circuit currentI ̄scis found from the circuit of Figure E3.1.8(c).

The KVL equations are given by

( 200 +j 20 )I ̄−j 20 I ̄sc= 20  0°

−j 20 I ̄+I ̄sc(j 20 −j 10 )=−

(

−j 100 I ̄

)

Solving forI ̄sc, one gets

I ̄sc= 12 I ̄=^240

200 −j 220

= 0. 807  47 .7° A

The Thévenin impedance at terminalsA–Bis then

Z ̄Th=

V ̄oc

I ̄sc

=

11. 94  84 .3°

0. 807  47 .7°

= 14. 8  36 .6°= 11. 88 +j 8. 82 

The Thévenin equivalent circuit at theA–Bterminals is shown in Figure E3.1.8(d).

(b) The impedance to be connected at terminalsA–Bfor matching is given by

Z ̄=Z ̄∗Th= 14. 8  − 36 .6°=RTh−jXTh= 11. 88 −j 8. 82 

(c) The maximum power transfer to the matched impedanceZ ̄is given by

Pmax=

(

Voc

2 RTh

) 2

RTh=

Voc^2

4 RTh

=

11. 942

4 × 11. 88

= 3 .0W

EXAMPLE 3.1.9

A single-phase source delivers 100 kW to a load operating at 0.8 lagging power factor. In order

to improve the system’s efficiency, power factor improvement (correction) is achieved through

connecting a capacitor in parallel with the load. Calculate the reactive power to be delivered by

the capacitor (considered ideal) in order to raise the source power factor to 0.95 lagging, and draw

the power triangles. Also find the value of the capacitance, if the source voltage and frequency

are 100 V (rms) and 60 Hz, respectively. Assume the source voltage to be a constant and neglect

the source impedance as well as the line impedance between source and load.